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6. (a) Which of these is a vector? A) energy B) force C) mass D) work (b) (i) State the equation that relates acceleration to change in velocity and time taken - Edexcel - GCSE Physics Combined Science - Question 6 - 2020 - Paper 1

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6.-(a)-Which-of-these-is-a-vector?----A)-energy----B)-force----C)-mass----D)-work--(b)-(i)-State-the-equation-that-relates-acceleration-to-change-in-velocity-and-time-taken-Edexcel-GCSE Physics Combined Science-Question 6-2020-Paper 1.png

6. (a) Which of these is a vector? A) energy B) force C) mass D) work (b) (i) State the equation that relates acceleration to change in velocity and tim... show full transcript

Worked Solution & Example Answer:6. (a) Which of these is a vector? A) energy B) force C) mass D) work (b) (i) State the equation that relates acceleration to change in velocity and time taken - Edexcel - GCSE Physics Combined Science - Question 6 - 2020 - Paper 1

Step 1

Which of these is a vector?

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Answer

The correct answer is B) force, as it has both magnitude and direction.

Step 2

State the equation that relates acceleration to change in velocity and time taken.

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Answer

The equation that relates acceleration to change in velocity and time is given by:

a=ΔvΔta = \frac{\Delta v}{\Delta t} where Δv \Delta v represents the change in velocity and Δt \Delta t represents the time taken.

Step 3

A van accelerates from a velocity of 2 m/s to a velocity of 20 m/s in 12 s. Calculate the acceleration of the van.

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Answer

To calculate the acceleration of the van, we first determine the change in velocity:

Δv=vfvi=20m/s2m/s=18m/s\Delta v = v_f - v_i = 20 \, \text{m/s} - 2 \, \text{m/s} = 18 \, \text{m/s}

Now, using the equation for acceleration:

a=ΔvΔt=18m/s12s=1.5m/s2a = \frac{\Delta v}{\Delta t} = \frac{18 \, \text{m/s}}{12 \, \text{s}} = 1.5 \, \text{m/s}^2

Therefore, the acceleration of the van is 1.5m/s21.5 \, \text{m/s}^2.

Step 4

Calculate the distance the cyclist travels in the 15 s.

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Answer

The distance traveled can be calculated by finding the area under the velocity-time graph. The area under the graph represents the distance covered, which can be calculated as:

Distance=Area=12×base×height=12×15s×7m/s=52.5m\text{Distance} = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \, \text{s} \times 7 \, \text{m/s} = 52.5 \, \text{m}

Thus, the cyclist travels a distance of 52.5m52.5 \, \text{m} in the 15 s.

Step 5

Explain how the stopping distance of a car is affected by factors involving the driver.

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Answer

The stopping distance of a car is influenced significantly by driver factors, including:

  • Reaction Time: A driver’s reaction time can vary based on alertness. A distracted or tired driver may take longer to react to a hazard, increasing the stopping distance.
  • State of Mind: If a driver is under the influence of drugs or alcohol, their judgment and reflexes are impaired, thereby increasing stopping distance.

For instance, a driver who is tired may not react timely to a red light, leading to an extended stopping distance.

Step 6

Explain how the stopping distance of a car is affected by factors involving the car or the road.

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Answer

Factors related to the car or the road that affect stopping distance include:

  • Tires: Worn or under-inflated tires can decrease traction, causing longer stopping distances.
  • Brakes: The condition of the brake system impacts stopping capability. Worn brakes may not apply adequate force.
  • Road Conditions: Wet or icy roads decrease friction between the tires and the surface, leading to increased stopping distances.

For example, if a car is on a slick road, its stopping distance can be significantly greater than on a dry road.

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